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Partial Implicit Differentiation

  • Thread starter mattmns
  • Start date
1,062
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Just wondering if I did this right:

Here is the question: find [tex]\frac{\partial z}{\partial x} of \frac{x^2}{9} - \frac{y^2}{4} + \frac{z^2}{2} = 1[/tex]

Now I put the [tex] \frac{\partial z}{\partial x} [/tex] on both sides then got.

[tex] \frac{2x}{9} - 0 + z \frac{\partial z}{\partial x} = 0 [/tex]

So


[tex] \frac{\partial z}{\partial x} = -\frac{2x}{9z} [/tex]

Now I know this is the right answer I am just curious if I did it right, it has been a while.

Now my reasoning:

[tex] \frac{\partial z}{\partial x} \frac{x^2}{9} [/tex] is just the derivative with respect to x, so it will be [tex]\frac{2x}{9}[/tex]

[tex] \frac{\partial z}{\partial x} \frac{y^2}{4} [/tex] has no z or x, so it is constand and therefore 0.

[tex] \frac{\partial z}{\partial x} \frac{z^2}{2} [/tex] has a z, so it is the derivative of itself, but times [tex] \frac{\partial z}{\partial x} [/tex]

Is all of that correct, or did I do something wrong. The book I have only shows one example :mad:

Thanks!
 
Last edited:

dextercioby

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It looks okay.You could have done it using the theorem of implicit functions,as well.

Daniel.
 
1,062
6
Thanks dex! I have no idea what the theorem of implicit functions is, but I probably could not use it anyway.
 
1,062
6
One more question for now :smile:

I need to find

[tex] \frac{\partial z}{\partial x} sin(xz) [/tex]

How exactly does this work?

Will it be: cos(xz) [times] what? [tex] \frac{\partial_}{\partial x} xz [/tex]

Which will then be cos(xz) [times] x ? This one seems to be wrong. Any ideas?
 

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